Problem: We know that $\frac{1}{3+\frac{1}{3}x^2}=\frac{1}{3}-\frac{1}{27}x^2+\frac{1}{243}{{x}^{4}}-\frac{1}{2187}{{x}^{6}}+...$ for $x\in\left(-3,3\right)$. Using this fact, find the function that corresponds to the following series. $ \frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\arctan(3x)$ (Choice B) B $\arctan(-3x)$ (Choice C) C $-\arctan(3x)$ (Choice D) D $\arctan\left(\frac{1}{3}x\right)$ (Choice E) E $-\arctan\left(\frac{1}{3}x\right)$ (Choice F) F $\arctan\left(-\frac{1}{3}x\right)$
Answer: First, note that the derivative of $ \frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ is $\frac{1}{3}-\frac{1}{27}x^2+\frac{1}{243}{{x}^{4}}-\frac{1}{2187}{{x}^{6}}+...=\frac{1}{3+\frac{1}{3}x^2}$ $\int\Big({\frac{1}{3}-\frac{1}{27}x^2+\frac{1}{243}{{x}^{4}}-\frac{1}{2187}{{x}^{6}}+...}\Big)dx=\frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ Then $\int{\frac{1}{3+\frac{1}{3}x^2}}~dx=\int\Big({\frac{1}{3}-\frac{1}{27}x^2+\frac{1}{243}{{x}^{4}}-\frac{1}{2187}{{x}^{6}}+...}\Big)dx$ $\int{\frac{\frac{1}{3}}{1+\frac{1}{9}x^2}}~dx=\frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ $\int{\frac{\frac{1}{3}}{1+\left(\frac{1}{3}x\right)^2}}~dx=\frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ $\arctan\left(\frac{1}{3}x\right) + C=\frac{x}{3}-\frac{{{x}^{3}}}{81}+\frac{{{x}^{5}}}{1215}-...$ Now let $x=0$ ; and since $\arctan(0)=0$, we know that $C=0$. Hence the given series is the power series for $\arctan\left(\dfrac{1}{3}x\right)$.